Given two immersions $f_i: M_i^m\to N_i^n $, $i=0,1$ we can always construct an immersion $f_0\# f_1 : M_0\#M_1\to N_0\#N_1$ restricting to $f_i$ over the image of $M_i\hookrightarrow M_0\#M_1$ in the connected sum. Indeed in a neighbourhood of a point $f_i$ behaves as an embedding $\mathbb D^m\to \mathbb D^n$ and by the uniqueness of such embedding up to isotopy we can find a diffeotopy $F:\mathbb D^m\times [0,1]\to \mathbb D^n\times [0,1] $ such that $F(\cdot, i) = f_i$ ($F(\cdot, t)$ doesn't have to send $\partial \mathbb D^m$ to $\partial \mathbb D^m$ for all $t$). Since $M_0\#M_1$ is constructed by gluing a tube $\mathbb D^m\times[0,1]$ to $M_0\bigsqcup M_1$ this gives $f_0\#f_1$.
Returning to your question, given $f:M^{m}\to \mathbb R^{2m}$, consider an immersion $g:\mathbb S^m\to \mathbb S^{2m}$ with a unique self intersection (this can be constructed by hand) and take the connected sum $f\# g: M\#\mathbb S^m\to \mathbb R^{2m}\#\mathbb S^{2m}$.